SWIFT 2 - alamofire - Parsing JSON parameters

Question

I am trying to call an object inside a JSON array using Alamofire. I can get Alamofire to parse my file however, I cannot find the way to get a specific parameter out.

if I use:

Alamofire.request(.GET, "https://example.com//users_result.json")
            .responseJSON { response in

                print(response.result)   // result of response serialization

                if let JSON = response.result.value {
                    print(JSON.count)
                    print("JSON: \(JSON[1])")
                }
        }

the code prints out:

> SUCCESS 4 JSON: {
>     Age = 22;
>     Email = "annahswift@gmail.com";
>     Location = "New York";
>     MyText = "Love Music";
>     Name = Annah;
>     Password = 123456;
>     REF = 003;
>     "Reg_Date" = "2015-07-30";
>     Sex = Female;
>     Surname = Swift; }

Now, how can I call the parameter 'name'?

print("JSON: \(JSON[1][4])") // nil ??

Possible duplicate of [How to parse JSON response from Alamofire API in Swift?](http://stackoverflow.com/questions/26114831/how-to-parse-json-response-from-alamofire-api-in-swift) (This one, or one of the thousands available.) — Eric Aya, Oct 27 '15 at 18:04
And also: [Swift documentation, The basics: Dictionaries](https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/CollectionTypes.html#//apple_ref/doc/uid/TP40014097-CH8-ID105) — Eric Aya, Oct 27 '15 at 18:06

score 0 · Answer 1 · answered Oct 27 '15 at 18:01

0

 print("Response data: \(JSON[1]["Name"] as! String)")

answered Oct 27 '15 at 18:01

SNos

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You have to escape the quotes, don't you? Or put it in separate statement. – Rob Oct 27 '15 at 18:11
2

@Rob Depends: we don't have to escape anymore since Xcode 7.1 beta 3. – Eric Aya Oct 27 '15 at 18:12

SWIFT 2 - alamofire - Parsing JSON parameters

1 Answers1